Figure 34-31 shows a coordinate system in front of a flat mirror, with the x-axis perpendicular to the mirror. Draw the image of the system in the mirror. (a) Which axis is reversed by the reflection? (b) If you face a mirror, is your image inverted (top for bottom)? (c) Is it reversed left and right (as commonly believed)? (d) What then is reversed?

Short Answer

Expert verified

(a) The x-axis is reversed by the reflection.

(b) If you face the mirror, then the image is not inverted.

(c) No, the image is not reversed left or right.

(d) Only the perpendicular i.e. x-axis is reversed.

Step by step solution

01

The given data:

A coordinate system placed in front of a mirror is given.

02

Understanding the concept of properties of the flat mirror:

According to the laws of reflection, the angle of reflection is equal to the angle of incidence. The image is obtained behind the plane which is present in the mirror. This process of obtaining a mirror image that is virtual and upright is known as a plane mirror reflection.

If you stood in front of a mirror, you would see your image. When the light rays coming from you hit the smooth surface of the mirror, they are reflected back at the same angle. Your eyes will see these reflected light rays as a mirror image.

When an item is reflected in a plane mirror, an image is created that is virtual, upright, facing left to right, and the same size as the thing.

03

(a) Calculation of the axis reversed by the reflection:

In the flat mirror reflection, the axis perpendicular to the mirror only gets reversed.

In this case, the x-axis is perpendicular.

The initial coordinates system is +x,+y,+z.

As you know, the image formed by a flat mirror is equally placed on the opposite side of the mirror with coordinates signs changing as per the concept of a coordinate system.

Again, as per the given figure, the mirror is placed on the axis perpendicular to it while in a parallel form to the y-axis and the z-axis.

Thus, after the reflection, the new coordinate system is given as follows:

(+x),+y,+z=x,+y,+z

Hence, the image distance is only reversed by the x-axis after the reflection.

04

(b) Calculation of the behavior after you face your mirror:

If you stand in front of the mirror, then your image formed will be exactly at a parallel distance to me on the opposite side of the mirror. Thus, only your image distance is equally opposite from the origin to the oppositex-axis but still remains unchanged in the y-axis image coordinate as the reflection is only reversed by the x-axis as per discussions in part (a).

Hence, there will be no change.

05

(c) Calculation for knowing if the image is reversed or not:

As per the question, you need to check the left and right sides of your image with the flat mirror. These directions indicate the coordinates of thez-axis that remains unchanged as per the discussions based on part (a).

Hence, there is no change in the left and right directions of the image.

06

(d) Calculation of the object that is reversed:

As per the detailed discussion in part (a), you can see that the image is reversed in the direction which is perpendicular to the mirror axis placement. As here, the mirror is placed perpendicular to thex-axis, thus the image is only inverted or reversed in that direction only.

Hence, only the perpendicular axis to the mirror is reversed i.e., the x-axis from calculations of part (a).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter


Isaac Newton, having convinced himself (erroneously as it turned out) that chromatic aberration is an inherent property of refracting telescopes, invented the reflecting telescope, shown schematically in Fig. 34-59. He presented his second model of this telescope, with a magnifying power of 38, to the Royal Society (of London), which still has it. In Fig. 34-59, incident light falls, closely parallel to the telescope axis, on the objective mirror. After reflection from the small mirror (the figure is not to scale), the rays form a real, inverted image in the focal plane (the plane perpendicular to the line of sight, at focal point F). This image is then viewed through an eyepiece. (a) Show that the angular magnification for the device is given by Eq. 34-15:

mθ=fob/fey

fob

the focal length of the objective is a mirror and

feyis that of the eyepiece.

(b) The 200 in. mirror in the reflecting telescope at Mt. Palomar in California has a focal length of 16.8 m. Estimate the size of the image formed by this mirror when the object is a meter stick 2.0 km away. Assume parallel incident rays. (c) The mirror of a different reflecting astronomical telescope has an effective radius of curvature of 10 m (“effective” because such mirrors are ground to a parabolic rather than a spherical shape, to eliminate spherical aberration defects). To give an angular magnification of 200, what must be the focal length of the eyepiece?

A point object is 10cmaway from a plane mirror, and the eye of an observer (with pupil diameter5.0mm) is 20cmaway. Assuming the eye and the object to be on the same line perpendicular to the mirror surface, find the area of the mirror used in observing the reflection of the point.

A grasshopper hops to a point on the central axis of a spherical mirror. The absolute magnitude of the mirror’s focal length is 40.0cm, and the lateral magnification of the image produced by the mirror is +0.200. (a) Is the mirror convex or concave? (b) How far from the mirror is the grasshopper?

58 through 67 61 59 Lenses with given radii. Objectstands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance, index of refraction n of the lens, radiusof the nearer lens surface, and radius of the farther lens surface. (All distances are in centimetres.) Find (a) the image distanceand (b) the lateral magnificationof the object, including signs. Also, determine whether the image is (c) realor virtual, (d) invertedfrom object or non-inverted, and (e) on the same side of the lens as objector on the opposite side.

Figure 34-47a shows the basic structure of the human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Fig. 34-47b). A “normal” eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Fig. 34-47c). (a) Suppose that for the parallel rays of Figs. 34-47a and b, the focal length fof the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40 cm, what focal length f of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free